Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 18}{x + 4} = \dfrac{-10x - 42}{x + 4}$
Multiply both sides by $x + 4$ $ \dfrac{x^2 - 18}{x + 4} (x + 4) = \dfrac{-10x - 42}{x + 4} (x + 4)$ $ x^2 - 18 = -10x - 42$ Subtract $-10x - 42$ from both sides: $ x^2 - 18 - (-10x - 42) = -10x - 42 - (-10x - 42)$ $ x^2 - 18 + 10x + 42 = 0$ $ x^2 + 24 + 10x = 0$ Factor the expression: $ (x + 4)(x + 6) = 0$ Therefore $x = -4$ or $x = -6$ At $x = -4$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -4$, it is an extraneous solution.